If the dimensions of a physical quantity are given by `M^(a)L^(b) T^(c)`, then the physical quantity will be

Options

A. velocity if a 1, b = 0, c =-1
B. acceleration if a = 1, b= 1, c = -2
C. force if a = 0, b = - 1, c = - 2
D. pressure if a = 1, b =-1, c = -2

Answer

D

Explanation

Pressure, `P= ("force")/("area")= ("mass" xx "acceleration")/("area")`
`:. [P]= ([M^(1)LT^(-2)])/([L^(2)])= [M^(1)L^(-1)T^(-2)]= M^(a)L^(b)T^(c)`
`:.` a=1, b =-1, c=-2